Question

If $xy={\tan }^{-1}\left(xy\right)+{\cot }^{-1}\left(xy\right),$ then $\frac{dy}{dx}$ is equal to

• A. $\frac{y}{x}$
• B. $\frac{-y}{x}$
• C. $\frac{x}{y}$
• D. $-\frac{x}{y}$

Answer 1

The correct option is B $\frac{-y}{x}$

$xy={\tan }^{-1}\left(xy\right)+{\cot }^{-1}\left(xy\right)$

Let $z=xy$

$y=\frac{z}{x}$

$\frac{dy}{dx}=\frac{x\frac{dz}{dx}-z\frac{dx}{dx}}{x^2}⟶\left(1\right)$

$z={\tan }^{-1}z+{\cot }^{-1}z$

$\frac{dz}{dx}=\frac{1}{1+z^2}\frac{dz}{dx}+\left(\frac{-1}{1+z^2}\right)\frac{dz}{dx}$

$\frac{dz}{dx}=0$

Putting value in $\left(1\right)$

$\frac{dy}{dx}=\frac{x\left(0\right)-z}{x^2}=-\frac{z}{x^2}=\frac{-xy}{x^2}=-\frac{y}{x}$

$\frac{dy}{dx}=-\frac{y}{x}$ [Answer B]